I am currently reading economics at Bristol University. I have always thoroughly enjoyed mathematics and hope to be able to instill that same enthusiasm in you too.
I am a patient person. I have coached hockey for children aged 5 to 16 and so have good experience in mentoring people.
I look forward to meeting you!
I am currently reading economics at Bristol University. I have always thoroughly enjoyed mathematics and hope to be able to instill that same enthusiasm in you too.
I am a patient person. I have coached hockey for children aged 5 to 16 and so have good experience in mentoring people.
I look forward to meeting you!
No DBS Check
To solve this quadratic equation we should first see if it is possible to factorise it into two seperate brakets. In this case there are no numbers which multiply to give -5 and add to give 3. Thus we should resort to using the quadratic formula which is used when equations cannot be solved by inspection. The quadratic formula is x=(-b+-(b2-4ac)1/2)/2a. Values for a,b and c can be lifted from the question. ax2+b-c=0 so in this question a=3 b=1 and c=-5. Inputing these into the equation will give two results. The answers are x=1.135 or x=-1.468 both of these answers make the original equation true.
To solve this quadratic equation we should first see if it is possible to factorise it into two seperate brakets. In this case there are no numbers which multiply to give -5 and add to give 3. Thus we should resort to using the quadratic formula which is used when equations cannot be solved by inspection. The quadratic formula is x=(-b+-(b2-4ac)1/2)/2a. Values for a,b and c can be lifted from the question. ax2+b-c=0 so in this question a=3 b=1 and c=-5. Inputing these into the equation will give two results. The answers are x=1.135 or x=-1.468 both of these answers make the original equation true.